From here:
It looks like:
It is supposed to say 'different from six' rather than 'different from three'
$T = \inf\{m: X_{m} = X_{m+1} = X_{m+2} = 6\}$
In every triple $P(all \ 6) = 1/216$
Thus, $P(\text{at least one is not 6}) = 215/216$
- By independence we have $P(T=k) = (215/216)^{\frac{k-1}{3}} \le (215/216)^{\frac{k-3}{3}}$:
1st triple: $(X_1, X_2, X_3)$
2nd triple: $(X_4, X_5, X_6)$
$$\vdots$$
$\frac{k-1}{3}$th triple: $(X_{k-3}, X_{k-2}, X_{k-1})$
$\frac{k+2}{3}$th triple: $(X_{k}, X_{k+1}, X_{k+2})$
Is that right? I don't see why we need to use an upper bound. I think the problem is saying something like
$$P(T = k) = (\frac{215}{216})^m < (\frac{215}{216})^{\frac{k-1}{3}}?$$
I'm not quite that's right considering the base is in between 0 and 1.
English is not my mother language, so I am not sure about the wordings here - but the idea should means that there is at least 1 number from the three tuples is not equal to $6$.
I guess the question may define $T$ as $$T \stackrel{\text{def}}= \inf\{m: X_{m-2}=X_{m-1}=X_m=6\}$$
Trivial, true by independence.
Here is the key spirit of this example. Let $B_m$ be the event that there is at least one number not equal to $6$ in $(X_m, X_{m+1}, X_{m+2})$. Then the event
$$ \{T = k\} = \bigcap_{m=1}^{k-3} B_m \cap B_{k-2}^c$$
Note that consecutive $B_m$ has "overlapping" $X_i$ so they are dependent. So instead we consider
$$ \{T = k\} \subseteq \bigcap_{m=1}^{\lfloor\frac {k-3} {3}\rfloor} B_{3m}$$
Now the events in the intersection are mutually independent, and hence we have the result. $$ \mathbb{P} \{T = k\} \leq \left(\frac {215} {216}\right)^{\lfloor\frac {k-3} {3}\rfloor}$$